MHT CET · Maths · Application of Derivatives
The equation of tangent to the curve \(y^{2}=a x^{2}+b\) at point \((2,3)\) is \(y=4 x-5\), then the values of \(a\) and \(\bar{b}\) are
- A \(3,-5\)
- B \(6,-5\)
- C 6,15
- D \(6,-15\)
Answer & Solution
Correct Answer
(D) \(6,-15\)
Step-by-step Solution
Detailed explanation
Given, \(\quad y^{2}=a x^{2}+b^{2}\)
\(\therefore\)
\(2 y \frac{d y}{d x}=2 a x\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{a x}{y}\)
\(\therefore\) Slope at \((2,3)=\left(\frac{d y}{d x}\right)_{(2,3)}=\frac{2 a}{3}\)
But slope of given tangent is \(m=4\) \(\frac{2 a}{3}=4 \Rightarrow \bar{a}=6\)
Since, point \((2,3)\) lies on the curve so, it satisfies the equation of the curve
\(\therefore\)
\((3)^{2}=6(2)^{2}+b\)
\(\Rightarrow \quad b=-15\)
\(\therefore\)
\(2 y \frac{d y}{d x}=2 a x\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{a x}{y}\)
\(\therefore\) Slope at \((2,3)=\left(\frac{d y}{d x}\right)_{(2,3)}=\frac{2 a}{3}\)
But slope of given tangent is \(m=4\) \(\frac{2 a}{3}=4 \Rightarrow \bar{a}=6\)
Since, point \((2,3)\) lies on the curve so, it satisfies the equation of the curve
\(\therefore\)
\((3)^{2}=6(2)^{2}+b\)
\(\Rightarrow \quad b=-15\)
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